What is the projection of #<3,4,-1 ># onto #<-1,3,-6 >#?

1 Answer
JMicheli
Jul 12, 2016

The projection is #(-15/sqrt46 , 45/sqrt46 , -90/sqrt46)#.

Explanation:

Given two vectors #vecV# and #vecW#, the projection of #vecV# onto #vecW# is given by:

#(vecV * vecW)vecW/||vecW||#

The inner product gives the component of #VecV# in the direction of #VecW#, and the fraction performs the same function as multiplying this magnitude by a unit vector in the direction of #vecW#.

So plugging in given values:
#vecV = (3,4,-1)#
#vecW = (-1,3,-6)#
#||vecW|| = sqrt(1 + 9 + 36) = sqrt(46)#

#vecV*vecW = (3*-1) + (4*3) + (-1*-6)#
#= -3 + 12 + 6 = 15#

So plugging everything in, the projection is:
#(15*(-1,3,-6))/sqrt(46) = (-15/sqrt46 , 45/sqrt46 , -90/sqrt46)#

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